Chapter 11.2, Problem 89AYU

To determine

To explain: The strategy to solve a system of linear equations with help of matrices.

Answer to Problem 89AYU

The solution set of the system of equations with same number of equations and variables is given by, $X=P^{-1}Q$ , where, X represents the variable matrix, P represent the coefficient matrix corresponding to variables, and Q represents the matrix of constants given on right hand of the equation.

Explanation of Solution

Given information:

The system of linear equation.

For a given system of liner equation the solution set of the system of equations with same number of equations and variables is given by, $X=P^{-1}Q$ , where, X represents the variable matrix, P represent the coefficient matrix corresponding to variables, and Q represents the matrix of constants given on right hand of the equation.

Write the augmented matrix of the system convert in reduced echelon form or directly found its inverse and solve the system.

Consider the system of equations as,

  $\begin{array}{c}5x-2y=26\\ x+2y=-2\end{array}$

Now, in the above system of equation we have two equations and two variables x and y.

We construct a coefficient matrix P ,

  $P=\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]$

Secondly, the variable matrix X is given as,

  $X=\left[\begin{array}{c}x\\ y\end{array}\right]$

Thirdly, matrix of constants Q which are given on right hand of the equation is,

  $Q=\left[\begin{array}{c}26\\ -2\end{array}\right]$

Therefore, we can write the system as,

  $\begin{array}{c}PX=Q\\ \left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}26\\ -2\end{array}\right]\end{array}$

First, we find the inverse of the matrix, $P=\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]$ .

Let the inverse of the matrix be $P^{-1}=\left[\begin{array}{ll}a\hfill & c\hfill \\ b\hfill & d\hfill \end{array}\right]$ .

Using the relation, $P{P}^{-1}=I_2$ , we get,

  $\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]\left[\begin{array}{ll}a\hfill & c\hfill \\ b\hfill & d\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$

Now, multiplying the above two matrices, we get,

  $\left[\begin{array}{ll}5a-2b\hfill & 5c-2d\hfill \\ a+2b\hfill & c+2d\hfill \end{array}\right]=\left[\begin{array}{ll}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$

Now, we equate the corresponding elements of the two matrices. First we write the system of equations when equated to the first column,

  $\begin{array}{c}5a-2b=1\\ a+2b=0\end{array}$

Now, we write the system of equations when equated to the second column,

  $\begin{array}{c}5c-2d=0\\ c+2d=1\end{array}$

So, we get the two augmented matrices which are as follows:

  $\left[\begin{array}{lll}5\hfill & -2\hfill & 1\hfill \\ 1\hfill & 2\hfill & 0\hfill \end{array}\right]$ and $\left[\begin{array}{lll}5\hfill & -2\hfill & 0\hfill \\ 1\hfill & 2\hfill & 1\hfill \end{array}\right]$

Combing the above two matrices such that the identity matrix is on the right hand side, we get,

  $\left[\begin{array}{llll}5\hfill & -2\hfill & 1\hfill & 0\hfill \\ 1\hfill & 2\hfill & 0\hfill & 1\hfill \end{array}\right]$

Now, we apply the Gauss-Jordan elimination to find the inverse of $\left[\begin{array}{llll}5\hfill & -2\hfill & 1\hfill & 0\hfill \\ 1\hfill & 2\hfill & 0\hfill & 1\hfill \end{array}\right]$ ,

First we have to make the entry $a_{11}$ as 1, we perform the row transformation as $R_1\iff R_2$ .

  $\left[\begin{array}{llll}1\hfill & 2\hfill & 0\hfill & 1\hfill \\ 5\hfill & -2\hfill & 1\hfill & 0\hfill \end{array}\right]$

Next, we have to make the entry $a_{21}$ as 0, we perform the row transformation as $R_2\to R_2-5R_1$ .

  $\left[\begin{array}{llll}1\hfill & 2\hfill & 0\hfill & 1\hfill \\ 0\hfill & -12\hfill & 1\hfill & -5\hfill \end{array}\right]$

Next, we have to make the entry $a_{22}$ as 1, we perform the row transformation as $R_2\to -\frac{R_2}{12}$ .

  $\left[\begin{array}{llll}1\hfill & 2\hfill & 0\hfill & 1\hfill \\ 0\hfill & 1\hfill & \frac{-1}{12}\hfill & \frac{5}{12}\hfill \end{array}\right]$

Next, we have to make the entry $a_{12}$ as 0, we perform the row transformation as $R_1\to R_1-2R_2$ .

  $\left[\begin{array}{llll}1\hfill & 0\hfill & \frac{1}{6}\hfill & \frac{1}{6}\hfill \\ 0\hfill & 1\hfill & \frac{-1}{12}\hfill & \frac{5}{12}\hfill \end{array}\right]$

Thus, two system can be segregated as follows,

  $\left[\begin{array}{lll}\frac{1}{6}\hfill & \frac{1}{6}\hfill & 1\hfill \\ \frac{-1}{12}\hfill & \frac{5}{12}\hfill & 0\hfill \end{array}\right]$ and $\left[\begin{array}{lll}\frac{1}{6}\hfill & \frac{1}{6}\hfill & 0\hfill \\ \frac{-1}{12}\hfill & \frac{5}{12}\hfill & 1\hfill \end{array}\right]$

Therefore, the inverse of the matrix $P=\left[\begin{array}{ll}5\hfill & -2\hfill \\ 1\hfill & 2\hfill \end{array}\right]$ is $P^{{}^{-1}}=\left[\begin{array}{cc}\frac{1}{6}& \frac{1}{6}\\ \frac{-1}{12}& \frac{5}{12}\end{array}\right]$ .

Now, substituting the values of $P^{{}^{-1}}$ , Q , and X in the equation $X=P^{-1}Q$ ,

  $\begin{array}{c}\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}& \frac{1}{6}\\ \frac{-1}{12}& \frac{5}{12}\end{array}\right]\left[\begin{array}{c}26\\ -2\end{array}\right]\\ =\left[\begin{array}{c}\left(\frac{1}{6}\right)\left(26\right)+\left(\frac{1}{6}\right)\left(-2\right)\\ \left(\frac{-1}{12}\right)\left(26\right)+\left(\frac{5}{12}\right)\left(-2\right)\end{array}\right]\\ =\left[\begin{array}{c}\frac{24}{6}\\ \frac{-36}{12}\end{array}\right]\\ =\left[\begin{array}{c}4\\ -3\end{array}\right]\end{array}$

Therefore, the value of x is 4 and y is $-3$ for the system of equations.

Thus, the preferred method is reduced echelon form.